Visited (Atomic masses: iron-56 = 55.9349 amu; hydrogen-1 = 1.0078 amu; mass of neutron = 1.0087 amu) HOW DO WE GET THERE? We must first calculate the mass defect (Am) for 5bFe. Since atomic masses (which include the electrons) are given, we must decide how to account for the electron mass: 55.9349 = mass of Fe atom = mass of nucleus + 26 m. 1.0078 = mass of H atom = mass of nucleus + me Thus, since a 56Fe nucleus is "synthesized" from 26 protons and 30 neutrons, we see that Am = (55.9349 - 26 m) - [26(1.0078 - m) + 30(1.0087)]. Solve for Am. amu

Visited (Atomic masses: iron-56 = 55.9349 amu; hydrogen-1 = 1.0078 amu; mass of neutron = 1.0087 amu) HOW DO WE GET THERE? We must first calculate the mass defect (Am) for 5bFe. Since atomic masses (which include the electrons) are given, we must decide how to account for the electron mass: 55.9349 = mass of Fe atom = mass of nucleus + 26 m. 1.0078 = mass of H atom = mass of nucleus + me Thus, since a 56Fe nucleus is "synthesized" from 26 protons and 30 neutrons, we see that Am = (55.9349 - 26 m) - [26(1.0078 - m) + 30(1.0087)]. Solve for Am. amu

Chemistry for Today: General, Organic, and Biochemistry

9th Edition

ISBN:9781305960060

Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. Hansen

Publisher:Spencer L. Seager, Michael R. Slabaugh, Maren S. Hansen

Chapter10: Radioactivity And Nuclear Processes

Section: Chapter Questions

Problem 10.8E

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Similar questions

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